What is actually Normalized Frequency?

I mean how could we define or describes it? I am new to MATLAB. I have calculated the the Frequency of 1st zero, Level of 1st slobe, Level of 2nd slobe and level of 3rd slobe in Rectangular, Triangular, Hann, Hamming, Blackman etc type in Frequency Domain after typing "wintool" in the window command of the MATLAB. Thanks

Answer 1

[Adam]

Normalised frequency is frequency in Hz (or more generically cycles/second or some other unit) divided by the sample frequency of your signal in Hz (or the same units as your original frequency).

So a normalised frequency of 1 represents your sampling frequency and 0.5 represents the Nyquist frequency.

[Tala Hed]

I think its, Fr/ * half* of sampling Fr.

[Kenny]

Could you please elaborate on your comment? What do you mean by you think it's ..... ? I also get confused about normalised frequency - but it's usually not because I don't know what normalising means. It's mainly because some documents don't appear to specify or say what the normalising factor is. Such as --- to normalise values based on the sampling frequency? Or normalise values based on 'half of the sampling frequency'? It's really the life story of many documents, tutorials etc. That is, lack of detail, or lost detail.

[Adam]

Normalising based on half frequency is not something I have ever seen and I'm not sure it makes much sense. Sure the values between 0.5 and 1 are not generally very useful but 0.5 is widely accepted as the Nyquist frequency and has a sensible interpretation as such - i.e. the frequency at Nyquist is 1 every 2 samples - e.g. the maximum frequency of peaks you can have in a signal is 1 every 2 because you have to have troughs between each peak, so 0.5, representing Nyquist, is the highest valid frequency for a signal

[Kenny]

Thanks Adam. For your original answer "So a normalised frequency of 1 represents your sampling frequency and 0.5 represents the Nyquist frequency." .....

The normalised value of 0.5 represents the case where a sinusoidal waveform having frequency 'f' is being sampled with a sampling frequency equal to the Nyquist rate (ie. 2f), right? So 0.5 doesn't actually represent the Nyquist frequency as such, right?

[Adam]

Yes, it should represent the Nyquist frequency, which is half the sample frequency by definition. You can't have Nyquits being equal to the sampling rate.

The highest frequency you can possibly have in a signal is Nyquist, which is every other sample being a peak (maxima) and every sample between those being a trough (minima), assuming we don't have a degenerate signal. So the frequency of the peaks is 0.5 (every other sample), likewise the frequency of the troughs. So 0.5 here is the Nyquist frequency, a frequency of 1 would be every sample (i.e. the sample rate/frequency).

[Kenny]

Thanks Adam! I see what's happening here now. I incorrectly assumed that Nyquist Frequency is Nyquist Rate, which are not the same definition. This is thanks (or actually no thanks) to those people that created two different definitions based on very similar terms. Thanks for your help again. Greatly appreciated.

[Robbin Roddewig]

Slight correction. The highest frequency that can be accurately represented is LESS THAN (not equal to) Nyquist. In you example of sampling peaks and troughs consider sampling the zero crossings if you are exactly half a wavelength based on your sampling of a sine wave being twice a wavelength. You would have zeros. Always. You cannot reconstruct a sine wave from zeros.

Answer 2

[George Fodor]

Normalization means to have a measure for a signal in the same, fixed, easy to use range such as [0,..,1] and this measure should not have physical units.

For normalizing the frequency f to a value fn in range [0,...,1], the most obvious way is to divide f by the sampling frequency fs:

fn = f / fs (1)

Formula (1) is a number (no units) showing how many sampling periods with frequency fs are in the sampled signal f.

However, according to Nyquist-Shannon theorem, the sampling frequency is always at least two times the frequency f. Thus (1) is never larger than 1/2. To have fn in the range [0, 1], we multiply (1) with a factor 2:

fn = 2 * f / fs (2)

Now if we replace fs to be the Nyquist frequency, fs=2*f then we obtain fn=1. For higher sampling frequencies than Nyquist frequency, the value fn will be less than one since we divide 2*f by an increasingly large number, thus we obtain the range [0,..,1] for fn as we wanted. If one would not use the multiplier 2 in (2), then the range would be always [0,...,1/2] which is not so nice.

In graphical representations in Matlab, the normalized frequency is shown as “(x pi rad/sample)”. This comes from the following calculation. It is known that omega in [rad/sec] and f in [Hz] are related by the formula:

omega = 2 * pi * f (3)

It means

f = omega / (2*pi) (4)

Replacing f above in (2) gives:

fn = 2*omega/(2*pi) / fs = (omega / pi) / fs (5)

We multiply (5) by pi and obtain:

fn * pi = omega / fs (6)

Now (omega / fs) has units [rad/sec] / [sample / sec] = [rad/sample]

Therefore we have fn in units [pi * rad / sample] as shown in graphics. However, creating a unit in this way is somewhat misleading since the relative frequency has no units, as seen in formula (1).

[Kenny]

Thanks very much George for your effort and time for explaining the normalisation. That was extremely useful. And I strongly believe that the beginning of every matlab tutorial on signal processing should show what you just wrote. That was excellent.

[Kenny]

Hello again George! For eqn (6), the updated normalised frequency could be labelled as something arbitrary - such as fn1, right? So fn1 = pi*fn, where fn is the value obtained from eqn (5). So the final "units" of fn1 is still rad/sample (since a multiplication of pi doesn't change the units, although the multiplication of pi is a scaling factor), right? Thanks George!

Another thing is ... if we normalise angular frequency --- such as:

wn = w/ws gives one normalised form.

then wn1 = 2w/ws gives another one.

Then wn1 = 2w/(2.pi.fs) = w/(pi.fs) --- yet another normalised form.

And finally wn2 = pi.wn1 = omega/fs (which is equation 6).

[Qiang Yang]

Thanks, it is very helpful!

[Zhao Lu]

Appreciate.

[Robbin Roddewig]

Hate to be picky but the sampling theorem is not twice the highest frequency but more than twice. If you sampled a sine wave twice per cycle you could get a series of zeros (sampling always at the zero crossings). With nothing but zeros you have an infinite number of sine waves (including no sine wave) that would fit those samples. The Nyquist frequency is 1/2 the sample rate but the Nyquist freqeuncy is not the highest frequency that can be accurately represented with that sampling rate.

[Jan]

I agree with @Kenny's first comment that a MATLAB course on signal processing should start with @George Fodor's statement. Perhaps every help page that uses the normalized frequency should have a fold out window (hidden by default) as a reminder. Especially, since the script

b = fir1(48,[0.35 0.65]);
freqz(b,1,512)

plots the magnitde and phase of the corresponding transfer function on a normalized frequency axis spanning [0,1), while the code

b = fir1(48,[0.35 0.65]);
[h, w] = freqz(b,1,512);

produces a radian frequency vector w between [0, pi). Indeed @Robbin Roddewig, pi is not included.

The help page that includes the above code (fir1) describes a bandpass filter based on a Hamming window with the 'normalized cutoff frequencies' 0.35 and 0.65, which are both neccesarily below Nyquist. Clearly the 1 in the plot and the pi in the w-vector both represent the Nyquist frequency, but the description should really be clearer in that regard.

[Sherman Chen]

Sorry but according to (6), should the unit of fn be rad/sample/pi, or rad/(sample*pi)? How come it is pi*rad/sample?